Intuitionistic Logic Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  ILE Home  >  Th. List  >  elabg GIF version

Theorem elabg 2688
 Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of [Quine] p. 44. (Contributed by NM, 14-Apr-1995.)
Hypothesis
Ref Expression
elabg.1 (𝑥 = 𝐴 → (𝜑𝜓))
Assertion
Ref Expression
elabg (𝐴𝑉 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
Distinct variable groups:   𝜓,𝑥   𝑥,𝐴
Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem elabg
StepHypRef Expression
1 nfcv 2178 . 2 𝑥𝐴
2 nfv 1421 . 2 𝑥𝜓
3 elabg.1 . 2 (𝑥 = 𝐴 → (𝜑𝜓))
41, 2, 3elabgf 2685 1 (𝐴𝑉 → (𝐴 ∈ {𝑥𝜑} ↔ 𝜓))
 Colors of variables: wff set class Syntax hints:   → wi 4   ↔ wb 98   = wceq 1243   ∈ wcel 1393  {cab 2026 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559 This theorem is referenced by:  elab2g  2689  intmin3  3642  finds  4323  elxpi  4361  ovelrn  5649  indpi  6440  peano5nnnn  6966  peano5nni  7917  peano5setOLD  10065
 Copyright terms: Public domain W3C validator